Abstract
It was recently proposed to encode the one-sided exponential source X via K parallel channels, Y1, ..., YK , such that the error signals X - Yi, i = 1,...,K, are one-sided exponential and mutually independent given X. Moreover, it was shown that the optimal estimator \hat{Y} of the source X with respect to the one-sided error criterion, is simply given by the maximum of the outputs, i.e., \hat{Y} = max{Y1,..., YK}. In this paper, we show that the distribution of the resulting estimation error X - \hat{Y} , is equivalent to that of the optimum noise in the backward test-channel of the one-sided exponential source, i.e., it is one-sided exponentially distributed and statistically independent of the joint output Y1,...,YK.
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